I was given this question by one of my friends:
Shape $ABCD$ is a parallelogram.
$AE = EB$
$BF = FC$
Find $\angle{FPC}$.
but when I tried to solve it, I found it impossible without at least a side value. I disagree with myself, as I am finding an angle, and size doesn't matter. I did this with trigonometry:
$$(EP) = \frac{(BC) \sin(180 - 110)}{\sin(90)}$$
This is using The Law of Sines. Then:
$$\angle{EFP} = \cos^{-1}(\frac{(EF)^2 + (FP)^2 - (EP)}{2(EF)(FP)})$$
This is using The Law of Cosines. Since $(EF) = (FP)$, I simplified that to:
$$\angle{EFP} = \cos^{-1}(\frac{2(EF)^2 - (EP)}{2(EF)^2}) = $$ $$\angle{EFP} = \cos^{-1}(1 - \frac{(EP)}{2(EF)^2})$$
From there, I went to:
$$\angle{FPC} = \frac{\angle{EFP}}2$$
Here I stopped. I thought something was wrong, so I went here to ask. Is there something I'm doing wrong? Did I make a mistake in my calculations? Is this question even possible? Note: please don't criticize me. I'm only grade 8.
There are two points I want to make.
(1) Your suggestion (about EF = PF) is correct. Your approach is correct too.
However, the proof can be slightly simplified by quoting the “intercept theorem”. It says:-
If BE // CP // FG, and F divides BC into two intercepts in the ratio BF : FC, then G will divide EP into two intercepts in the same ratio. That is, EG : GP = BF : FC.
Therefore, after constructing a line through F parallel to BE cutting EP at G, we have EG = GP because BF = FC.
Because $\angle FGP = \angle BEG = 90^0$, FG is actually the perpendicular bisector of EP. Thus, your argument about symmetry can continue.
(2) This problem has no solution.
The diagram above shows if we draw another parallelogram (AB’C’D) having the same characteristics as ABCD, $\angle FPC$ will not equal to $\angle F’P’C’$. This means that angle is NOT constant and is quite length (of $AB$, say) dependent.